To prove the derivative of the inverse sine function $\sin^{-1}(x)$, let's denote:

$y = \sin^{-1}(x)$

Since $y = \sin^{-1}(x)$, we have: $\sin(y) = x$

Our goal is to find $\frac{d}{dx}(\sin^{-1}(x)) = \frac{dy}{dx}$.

Step 1: Differentiate Both Sides

Differentiate both sides of $\sin(y) = x$ with respect to $x$:

$\frac{d}{dx}(\sin(y)) = \frac{d}{dx}(x)$

Using the chain rule on the left side, we get:

$\cos(y) \cdot \frac{dy}{dx} = 1$

Step 2: Solve for $\frac{dy}{dx}$

Now, isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\cos(y)}$

Step 3: Express $\cos(y)$ in Terms of $x$

Since $\sin(y) = x$, we can use the Pythagorean identity $\sin^2(y) + \cos^2(y) = 1$ to express $\cos(y)$ in terms of $x$:

$\cos^2(y) = 1 - \sin^2(y)$

Since $\sin(y) = x$, this becomes:

$\cos^2(y) = 1 - x^2$

Taking the square root of both sides (and assuming $\cos(y)$ is positive in the domain of $\sin^{-1}(x)$):

$\cos(y) = \sqrt{1 - x^2}$

Step 4: Substitute Back

Substitute $\cos(y) = \sqrt{1 - x^2}$ into our expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

Conclusion

Thus, the derivative of $\sin^{-1}(x)$ with respect to $x$ is:

$\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}$

This completes the proof.

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Would you like further details or have any questions? Here are five related questions you might find helpful:

1. How do you derive the derivative of the inverse cosine function?
2. Can you prove the derivative of the inverse tangent function?
3. How is the Pythagorean identity used in finding inverse trigonometric derivatives?
4. How does the chain rule apply when differentiating inverse functions?
5. How do we handle the domain restrictions for inverse trigonometric functions?

Tip: Always keep track of domain restrictions in inverse functions, as they affect the validity of results and square root simplifications.